Problem: Is ${103062}$ divisible by $3$ ?
Solution: A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {103062}= &&{1}\cdot100000+ \\&&{0}\cdot10000+ \\&&{3}\cdot1000+ \\&&{0}\cdot100+ \\&&{6}\cdot10+ \\&&{2}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {103062}= &&{1}(99999+1)+ \\&&{0}(9999+1)+ \\&&{3}(999+1)+ \\&&{0}(99+1)+ \\&&{6}(9+1)+ \\&&{2} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {103062}= &&\gray{1\cdot99999}+ \\&&\gray{0\cdot9999}+ \\&&\gray{3\cdot999}+ \\&&\gray{0\cdot99}+ \\&&\gray{6\cdot9}+ \\&& {1}+{0}+{3}+{0}+{6}+{2} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${103062}$ is divisible by $3$ if ${ 1}+{0}+{3}+{0}+{6}+{2}$ is divisible by $3$ Add the digits of ${103062}$ $ {1}+{0}+{3}+{0}+{6}+{2} = {12} $ If ${12}$ is divisible by $3$ , then ${103062}$ must also be divisible by $3$ ${12}$ is divisible by $3$, therefore ${103062}$ must also be divisible by $3$.